Question:
If the coefficients of $2 \mathrm{nd}, 3 \mathrm{rd}$ and 4 th terms in the expansion of $(1+x)^{2 \mathrm{n}}$ are in A.P., show that $2 n^{2}-9 n+7=0$.
Solution:
Given:
$(1+x)^{2 n}$
Thus, we have:
$T_{2}=T_{1+1}$
$={ }^{2 n} C_{1} x^{1}$
$T_{3}=T_{2+1}$
$={ }^{2 n} C_{2} x^{2}$
$T_{4}=T_{3+1}$
$={ }^{2 n} C_{3} x^{3}$
We have coefficients of the 2 nd, 3 rd and 4 th terms in AP.
$\therefore 2\left({ }^{2 n} C_{2}\right)={ }^{2 n} C_{1}+{ }^{2 n} C_{3}$
$\Rightarrow 2=\frac{{ }^{2 n} C_{1}}{{ }^{2 n} C_{2}}+\frac{{ }^{2 n} C_{3}}{{ }^{2 n} C_{2}}$
$\Rightarrow 2=\frac{2}{2 n-1}+\frac{2 n-2}{3}$
$\Rightarrow 12 n-6=6+4 n^{2}-4 n-2 n+2$
$\Rightarrow 4 n^{2}-18 n+14=0$
$\Rightarrow 2 n^{2}-9 n+7=0$
Hence proved.